for θ θ
− p
, then there is a NE path γ from 0 to r, θ with a passage time less than C
8
r. Therefore, by 4.13 and 4.14
P[~ T
0, r, θ ≤ C
8
r]
≤ P[∃ a NE γ from 0 to r, θ with more than C r closed edges,
these closed edges contain less than C r 2 bad edges, ~
T γ ≤ C
8
r] + C
9
exp −C
10
r
≤ P[∃ a NE γ from 0 to r, θ , γ contains less than C r2 bad edges, ~T γ ≤ C
8
r] +C
9
exp −C
10
r. 4.15
If there is a NE path from 0 to r,
θ with less than C r2 bad edges among these C r closed edges, note that each good edge costs at least passage time
δ, so the passage time of the path is more than δC r2. Thus, if we select C
8
such that C
8
Cδ2,
P[ ∃ a NE γ from 0 to r, θ , γ contains less than C r2 bad edges, ~T γ ≤ C
8
r] = 0. 4.16
By 4.15 and 4.16, for F 0 ≥ ~p
c
, θ θ
− p
, P[~
T 0, r,
θ ≤ C
8
r] ≤ C
9
exp −C
10
r. 4.17
When θ θ
+ p
, by symmetry, we still have 4.17. Therefore, Theorems 2 and 4 follow.
5 Inside the percolation cone.
In Section 5, we assume that F 0 = p ~p
c
and θ ∈ [θ
− p
, θ
+ p
]. Edge e is called an open or a closed edge if te = 0 or te
0, respectively. We define τe = 0 if te = 0, or τe = 1 if te 0. We also denote by ~
T
τ
u, v the passage time corresponding to τe. Let
B
τ
t = {v ∈ Z
2
: ~ T
τ
0, v ≤ t}. 5.1
We also assume that r cos θ , r sin θ ∈ Z
2
for r 0 and θ ∈ [θ
− p
, θ
+ p
] without loss of generality. If r,
θ ∈ B
τ
t, then ~
T
τ
0, r, θ ≤ t.
5.2 Note that B
τ
t will eventually cover all the vertices in R
+
× [θ
− p
, θ
+ p
] as t → ∞, so for any r, θ ∈
R
+
× [θ
− p
, θ
+ p
], there exists a t such that r, θ ∈ B
τ
t. Let σ be the smallest t such that r, θ ∈
B
τ
t. We will estimate σ to show that there exist positive constants C
i
= C
i
F for i = 1, 2 such that for all large k,
P[
σ ≥ k] ≤ C
1
exp −C
2
k uniformly in r 0 and θ ∈ [θ
− p
, θ
+ p
]. 5.3
Note that
P[
σ ≥ k] = X
Γ
P[
σ ≥ k, B
τ
k − 2 = Γ], where Γ, containing the origin, takes all possible vertex sets in the first quadrant. We also remark
that for distinct Γ
1
and Γ
2
, {σ ≥ k, B
τ
k − 2 = Γ
1
} and {σ ≥ k, B
τ
k − 2 = Γ
2
} are disjoint. 2280
If σ ≥ k and B
τ
k − 2 = Γ, then Γ does not contain r, θ : Γ ∩ r, θ = ;.
5.4 In other words, the above sum is over all Γ that do not contain r,
θ . Thus, by Lemma 5, there is no NE open path from
∂
o
Γ to r, θ . Otherwise, σ k, which is contrary to the assumption
that σ ≥ k. For a fixed Γ, we denote by E
k
Γ the above event that there is no NE open path from ∂
o
Γ to r, θ . Thus, there exists a NE open path outside Γ ∪ ∂
e
Γ from ∂
o
Γ to r, θ . Note that
E
k
Γ only depends on configurations of edges outside Γ ∪ ∂
e
Γ, so by Lemma 6, for any fixed Γ with Γ ∩ r, θ = ;,
E
k
Γ and {B
τ
k − 2 = Γ} are independent. 5.5
By 1.5, there exists 0 δ 1, uniformly in r 0 and θ ∈ [θ
− p
, θ
+ p
] such that for any fixed Γ that does not contain r,
θ ,
P[ E
k
Γ] ≤ 1 − P[0 → r, θ ] ≤ 1 − δ. 5.6
With these observations,
P[ σ ≥ k] =
X
Γ
P[ σ ≥ k, B
τ
k − 2 = Γ] ≤
X
Γ
P[B
τ
k − 2 = Γ, E
k
Γ] ≤
X
Γ
P[B
τ
k − 2 = Γ]1 − δ, where the sum is over all Γ containing the origin but not r,
θ . Note that for a fixed Γ containing the origin but not r,
θ , by Lemma 5 again, {B
τ
k − 2 = Γ} ⊂ {σ ≥ k − 2}. Therefore,
P[ σ ≥ k] ≤ 1 − δP[σ ≥ k − 2].
5.7 Thus, 5.3 follows if we iterate 5.7. We show Theorem 1 by 5.3. In fact, if te is bounded from
above by a constant, then Theorem 1 is implied by 5.3 directly. However, if we restrict ourselves only on a moment condition, the proof is more intricate:
Proof of Theorem 1. Given σ, the passage time T 0, r, θ is dominated by the sum of σ copies
of the weight distribution, conditioned to be non-zero. This is because, conditioned on the set of closed edges, the weights of the closed edges are i.i.d. and each has the original weight distribu-
tion conditioned on being non-zero. We simply choose an arbitrary path from the origin to r, θ
containing σ closed edges for example, choose the first one in the lexicographic order, and sum
the weights of the closed edges along that path to get an upper bound on the passage time. Let Y
1
, Y
2
, · · · represent i.i.d. copies of the edge-weight distribution conditioned to be non-zero. If the
original weight-distribution has a finite mth moment, then so does this distribution. Since σ is only
affected by zero-edges, {Y
i
} and σ are also independent. Thus, by 5.3, we get
E~ T
0, r, θ
m
= E
E~
T 0, r,
θ
m
| σ
≤ E EY
1
+ Y
2
+ · · · + Y
σ m
≤ E
σ
m
EY
m 1
=
E
σ
m
EY
m 1
∞. 2281
So Theorem 1 follows.
6 Critical phase.
Proof of Theorem 5. In Section 6, we assume that F 0 = ~p. There are two possible behaviors for
F at 0: a either there exists a small h such that F x = F 0 =
~p
c
for x ∈ [0, h],
b or there exists a sequence {x
n
} with x
n
↓ 0 such that Fx
n
↓ F0 and Fx
n
F 0. Let us assume that case b holds. For each n, we construct another distribution:
G
n
x = ¨
F 0 if 0
≤ x x
n
, F x
if x
n
≤ x. By this definition, for each n,
G
n
x
n
~p
c
. By 1.5, for all r, there exists a directed path from the origin to r,
π4 such that its passage time in each edge is at most x
n
with a positive probability. By 1.1, for each n, ~
µ
G
n
π4 ≤ 2x
n
. 6.1
By Lemma 7, ~
µ
F
π4 ≤ ~µ
G
n
π4 ≤ 2x
n
. 6.2
By 6.2, we can show that ~
µ
F
π4 = 0. 6.3
Therefore, Theorem 5 follows if case b holds. Now we focus on case a. Note that F cannot be flat forever, so there are points h
1
h 0 such that F h
1
F 0 and F x = F 0 for 0 ≤ x ≤ h. Now we assume that F satisfies the following extra condition:
i. There exists h 0 such that
F x = F 0 = ~p
c
, when 0 ≤ x h, and Fx ~p
c
, when x ≥ h.
In other words, there is a jump point at h. We focus on case a i. We take
ε 0 small such that F 0 +
ε F h. Then we construct another distribution:
G
ε
x = ¨
F 0 + ε
if 0 ≤ x h,
F x if h
≥ x. As we defined, te is the random variable with distribution F . Let g
ε
e be the random variable with distribution G
ε
. We couple te and g
ε
e and then assume that all the couples te, g
ε
e
e ∈L
2
are independent and identically distributed. Define g
ε
e as follows: 2282
If te = 0, then g
ε
e = 0. If te = x
h, then g
ε
e = x. If te = h, then
g
ε
e ¨
= 0 with probability
ε F h − ~p
c −1
, = h
with probability 1 − ε Fh − ~p
c −1
. We need to verify that g
ε
has distribution G
ε
. Since the verification is simple, we leave to the readers.
Now we show Theorem 5 under case a i. Let γ
t
be an optimal path for ~ T
t
0, r, π4 with time
Kategori